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Finding marginal PDFs

nacho

Active member
Sep 10, 2013
156
Please see the attached image for my question. I don't understand how to compute the integral, is there some trick?
I do believe that
to find fx(x) I integrate the joint pdf, with respect to x with the bounds set as the range of Y. But this leaves me with a very complex integration
Similarly for fy(y). Is there some trick?

Any help is greatly appreciated.
 

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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Please see the attached image for my question. I don't understand how to compute the integral, is there some trick?
I do believe that
to find fx(x) I integrate the joint pdf, with respect to x with the bounds set as the range of Y. But this leaves me with a very complex integration
Similarly for fy(y). Is there some trick?

Any help is greatly appreciated.

Hi nacho!

To find $f_X(x)$ you need to integrate $f_{X,Y}(x,y)$ with respect to y.

That is:
$$f_X(x) = \int_{-\infty}^{+\infty} f_{X,Y}(x,y) dy$$

The problem is of course that exponent, but you can complete the square, do a substitution, and use the standard integral of a normal distribution.

That is:
$$f_X(x) = \int_{-\infty}^{+\infty} f_{X,Y}(x,y) dy
= \int_{-\infty}^{+\infty} \frac{1}{\pi\sqrt 2} \exp\left(-(y + \frac 1 2 x \sqrt 2)^2 - \frac 1 2 x^2\right) dy$$

Can you substitute $w = y + \frac 1 2 x \sqrt 2$?

And use that \(\displaystyle \int_{-\infty}^{+\infty} \exp\left(-\frac 1 2 u^2\right) du = \sqrt{2\pi}\)?